Chapter 1.6, Problem 50E

To determine

To plot the resulting function of given discrete-time dynamical system and use this function to plot the concentration after 10 days.

Answer to Problem 50E

It is observed from the concentration values, that the concentration reaches the value 2 after 10 days.

Explanation of Solution

Given:

The discrete-time dynamical system :

  $M_{t+1}=\frac{1}{2}{M}_t+1$

Consider the following discrete-time dynamical system representing the medication

  $M_{t+1}=\frac{1}{2}{M}_t+1$

Let us find the concentration of medication after 10 days by following the above system 10 times

Let the initial concentration be $M_0$

Therefore, the concentration after 1 day is as follows

  $M_1=\frac{1}{2}{M}_0+1$ ....(1)

The concentration after 2 days is as follows

  $\begin{array}{l}{M}_2=\frac{1}{2}{M}_1+1\\ =\frac{1}{2}\left(\frac{1}{2}{M}_0+1\right)+1\\ =\frac{1}{2^2}{M}_0+1+\frac{1}{2}\end{array}$ .......(2)

The concentration after 3 days is as follows

  $\begin{array}{l}{M}_3=\frac{1}{2}{M}_2+1\\ =\frac{1}{2}\left(\frac{1}{2^2}{M}_0+1+\frac{1}{2}\right)+1\\ =\frac{1}{2^3}{M}_0+1+\frac{1}{2}+\frac{1}{2^2}\end{array}$ ...........(3)

Observing the pattern of concentration on first 3 days represented by equation (1), equation (2) and equation (3), the concentration on 10th day is as follows

  $M_{10}=\frac{1}{2^{10}}{M}_0+\left(1+\frac{1}{2}+\frac{1}{2^2}+\mathrm{....}\frac{1}{2^{10}}\right)$

  $=\frac{1}{2^{10}}{M}_0+\frac{1\left(1-{\left(\frac{1}{2}\right)}^{11}\right)}{1-\frac{1}{2}}$ ( Using the formula of geometric progression series)

  $\begin{array}{l}=\frac{1}{2^{10}}{M}_0+\frac{2\left(2^{11}-1\right)}{2^{11}}\\ =\frac{1}{2^{10}}{M}_0+\frac{2\left(2047\right)}{2048}\\ =\frac{1}{2^{10}}{M}_0+\frac{4094}{2048}\mathrm{...................}(4)\end{array}$

Graph of the function is shown below

  

Let us find the concentration after 10 days if the starting concentration is 1.0

  $\begin{array}{l}{M}_{10}=\frac{1}{2^{10}}{M}_0+\frac{4094}{2048}\mathrm{...................}(\text{From}(4))\\ =\frac{1}{2^{10}}(1.0)+\frac{4094}{2048}\\ =2\end{array}$

Let us find the concentration after 10 days if the starting concentration is 5.0

  $\begin{array}{l}{M}_{10}=\frac{1}{2^{10}}{M}_0+\frac{4094}{2048}\mathrm{...................}(\text{From}(4))\\ =\frac{1}{2^{10}}(5.0)+\frac{4094}{2048}\\ =2.0039\end{array}$

Let us find the concentration after 10 days if the starting concentration is 18.0

  $\begin{array}{l}{M}_{10}=\frac{1}{2^{10}}{M}_0+\frac{4094}{2048}\mathrm{...................}(\text{From}(4))\\ =\frac{1}{2^{10}}(18.0)+\frac{4094}{2048}\\ =2.0166\end{array}$

It is observed from the concentration values, that the concentration reaches the value 2 after 10 days.

Thus, to achieve the goal of stable concentration of 2.0 miligrams per liter, this therapy is good.